Giải thích các bước giải:
Ta có:
$\cos^4x-\cos^4(\dfrac{\pi}{2}-x)$
$=\cos^4x-\sin^4(x)$
$=(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)$
$=(\cos^2x-\sin^2x)\cdot 1$
$=\cos^2x-\sin^2x$
$=2\cos^2x-(\cos^2x+\sin^2x)$
$=2\cos^2x-1$
$=2(-\cos(\pi+x))^2-1$
$=2\cos^2(\pi+x)-1$